Answer:
The value is 
Explanation:
From the question we are told that
The rate at which the plate rotates is 
Generally the revolution per second is mathematically represented as

=> 
Sure. Body can move with uniform speed, and having zero velocity, when velocity becomes zero due to change in direction over time t.
For Example. - An Object is moving with uniform speed in a circular path, then after one complete revolution, it's velocity is zero, but speed still remains uniform
Hope this helps!
Answer:
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Explanation:
This is an angular kinematic exercise the equation for the angular position
the particle A
θ = θ₀ + ω₀ t + ½ α t²
They say for the particle B
w₀B = ½ w₀
αB = 2 α
In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial
t´ = t - t_1
l
et's write the equation of particle B
θ = θ₀ + w₀B t´ + ½ αB t´2
replace
θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Answer:
4.15 m/s
Explanation:
Its given that acceleration is 0.1 m/s² with a direction opposite to the velocity. Since, the direction of acceleration is opposite to the velocity, this gives us a hint that the velocity is decreasing and so acceleration would be negative.
i.e.
acceleration = a = - 0.1 m/s²
Distance covered = S = 6m
Velocity after covering 6 meters = Final velocity =
= 4 m/s
We need to find the initial speed, which will be the same as the magnitude of initial velocity.
Initial velocity =
= ?
3rd equation of motion relates the acceleration, distance, final velocity and initial velocity as:

Using the known values in the formula, we get:

Thus, the initial speed of the ball was 4.15 m/s
We can find the force by using the following formula;
N = ma + mg
Fa = ma = 76 x 1.2 = 91.2
Fg = mg = 76 x 9.8 = 744.8
N = 91.2 + 744.8 = 836
So, the force is 836 N.